Hi I'm new to this, I've been given a question on compound angles and I really don't know how to go about it. Any help would be greatly appreciated:
Using compound angles prove that Cos(y-pi)+Sin(y+pi/2)=0
thanks Shaun
Hello, Shaun!
You are expected to know these Compound Angle Identities:
. . $\displaystyle \sin(A \pm B) \:=\:\sin A\cos B \pm \cos A\sin B$
. . $\displaystyle \cos(A \pm B) \:=\:\cos A\cos B \mp \sin A\sin V$
Using compound angles prove that: .$\displaystyle \cos(x-\pi)+\sin(x+\tfrac{\pi}{2})\:=\:0$
We have: .$\displaystyle \cos(x - \pi) \:=\:\cos(x)\cos(\pi) + \sin(x)\sin(\pi)$
. . . . . . . . . . . . . . . . $\displaystyle =\;\cos(x)\!\cdot\!(\text{-}1) + \sin(x)\!\cdot\!(0) $
. . . . . . . . . . . . . . . . $\displaystyle =\;\text{-}\cos(x)$
We have: .$\displaystyle \sin(x +\tfrac{\pi}{2}) \;=\;\sin(x)\cos(\tfrac{\pi}{2}) + \cos(x)\sin(\tfrac{\pi}{2}) $
. . . . . . . . . . . . . . . . $\displaystyle =\;\sin(x)\!\cdot\!(0) + \cos(x)\!\cdot\!(1)$
. . . . . . . . . . . . . . . . $\displaystyle =\;\cos(x)$
Therefore: .$\displaystyle \cos(x-\pi) + \sin(x + \tfrac{\pi}{2}) \;=\;\text{-}\cos(x) + \cos(x) \;=\;0$